proof of Hadamard’s inequality

Let’s first prove the second inequality. If A is singular, the thesis is trivially verified, since for a Hermitian positive semidefinite matrix the right-hand side is always nonnegative, all the diagonalMathworldPlanetmath entries being nonnegative. Let’s thus assume det(A)0, which means, A being Hermitian positive semidefinitePlanetmathPlanetmath, det(A)>0. Then no diagonal entry of A can be 0 (otherwise, since for a Hermitian positive semidefinite matrix, 0λminaiiλmax, λmin and λmax being respectively the minimal and the maximal eigenvalueMathworldPlanetmathPlanetmathPlanetmathPlanetmath, this would imply λmin=0, that is A is singular); for this reason we can define D=diag(d11,d22,,dnn), with dii=aii-12, since all aii+. Let’s furthermore define B=DAD. It’s easy to check that B too is Hermitian positive semidefinite, so its eigenvalues λB are all non-negative (actually, since A=AH and since D is real and diagonal, BH=(DAD)H=DH(DA)H=DHAHDH=DAD=B; on the other hand, for any 𝐱𝟎, 𝐱HB𝐱=𝐱HDAD𝐱=(𝐱HD)A(D𝐱)=(DH𝐱)HA(D𝐱)=(D𝐱)HA(D𝐱)=𝐲HA𝐲0). Moreover, we have obviously bii=diiaiidii=1 so that tr(B)=n and, recalling the geometric-arithmetic mean inequality (, which holds in this case because the eigenvalues of B are all non-negative,


and since det(B)=det(D)2det(A)=(i=1naii)-1det(A), we have the thesis. Since det(A)=i=1naii if and only if det(B)=1 and since in the geometric-arithmetic inequality equality holds if and only if all terms are equal, we must have λB(i)=λB, so that i=1nλB(i)=λBn=1, whence λB=1 (λB having to be non-negative), and since B is Hermitian and hence is diagonalizable, we obtain B=I, and so A=D-1BD-1=D-2=diag(a11,a22,,ann). So we can conclude that equality holds if and only if A is diagonal.

Let’s now derive the more general first inequality. Let A be a complex-valued n×n matrix. If A is singular, the thesis is trivially verified. Let’s thus assume det(A)0; then B=AAH is a Hermitian positive semidefinite matrix (actually, BH=(AAH)H=(AH)HAH=AAH=B and, for any 𝐱𝟎,𝐱HB𝐱=𝐱HAAH𝐱=(𝐱HA)(AH𝐱)=(AH𝐱)H(AH𝐱)=𝐲H𝐲=𝐲220). Therefore, the second inequality can be applied to B, yielding:


As we proved above, for det(B) to be equal to i=1nbii, B must be diagonal, which means that k=1naikajk=|aii|2δij. So we can conclude that equality holds if and only if the rows of A are orthogonalMathworldPlanetmathPlanetmath.


  • 1 R. A. Horn, C. R. Johnson, Matrix Analysis, Cambridge University Press, 1985
Title proof of Hadamard’s inequality
Canonical name ProofOfHadamardsInequality
Date of creation 2013-03-22 15:37:02
Last modified on 2013-03-22 15:37:02
Owner Andrea Ambrosio (7332)
Last modified by Andrea Ambrosio (7332)
Numerical id 17
Author Andrea Ambrosio (7332)
Entry type Proof
Classification msc 15A45